Question

An airplane has an airspeed of 400 kilometers per hour bearing N45oE. The wind velocity is 50 kilometers per hour in the direction N30oW.  Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is its direction?

Answer 1

see the figure below to better understand the problem

step 1

Find out the resultant vector in the x-coordinate

$R_x=400\cdot\cos 45^o-50\cdot\cos 30^o$$R_x=400\cdot\frac{\sqrt[]{2}}{2}-50\cdot\frac{\sqrt[]{3}}{2}$$\begin{gathered} R_x=200\sqrt[]{2}-25\sqrt[]{3} \\ R_x=239.5 \end{gathered}$

step 2

Find out the resultant in the y-coordinate

$R_y=400\cdot\sin 45^o+50\cdot\sin 30^o$$R_y=400\cdot\frac{\sqrt[]{2}}{2}+50\cdot\frac{1}{2}$$\begin{gathered} R_y=200\cdot\sqrt[]{2}+25 \\ R_y=307.8 \end{gathered}$

step 3

Find out the angle of the resultant

$\tan \theta=\frac{R_y}{R_x}$$\tan \theta=\frac{200\cdot\sqrt[]{2}+25}{200\sqrt[]{2}-25\sqrt[]{3}}$

simplify

$\tan \theta=\frac{25\lbrack8\cdot\sqrt[]{2}+1\rbrack}{25\lbrack8\sqrt[]{2}-\sqrt[]{3\rbrack}}$$\begin{gathered} \tan \theta=\frac{8\cdot\sqrt[]{2}+1}{8\sqrt[]{2}-\sqrt[]{3}} \\ \theta=52.1^o \end{gathered}$

Find out the resultant vector

$R=\sqrt[]{Ry^2+Rx^2}^{}_{}$$\begin{gathered} R=\sqrt[]{307.8^2+239.5^2} \\ R=390\text{ }\frac{km}{h} \end{gathered}$

therefore

The resultant vector is 390 km per hour N52.1 E